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3 years ago

Which of the following is true of a heterogeneous mixture?

2 answers:

8 0

A, a heterogeneous mixture is a mixture where all the particles are not made up of one uniform composition. you can distinguish the different particles

Leno4ka [110]3 years ago

3 0

Answer: Option (a) is the correct answer.

Explanation:

A heterogeneous mixture is defined as the mixture in which distribution of solvent particles is uneven throughout the solution.

For example, a mixture of sand and water is a heterogeneous mixture.

So, when we take different samples for the same heterogeneous mixture then there will be different composition of solute particles in each of them.

On the other hand, in a homogeneous solution there will be even distribution of solute particles into the solvent. Each sample of a homogeneous mixture will contain same composition of solute particles.

Thus, we can conclude that the statement different samples may not be exactly alike, is true of a heterogeneous mixture.

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43 moles feci3 are dissolved in 0.64l of solution what is the molarity of the solution

nasty-shy [4]

Molarity=moles/liter
molarity=43/0.64
molarity=67.19moles/litre

4 0

3 years ago

An ideal gas has a volume of 14.5 L in a flexible balloon at 680.0 mmHg at

irina1246 [14]

Answer:a

Explanation:yes

6 0

3 years ago

If molar mass of M(OH)3 = 78 8. mass of M

Cerrena [4.2K]

Answer:

From molar mass=total RAM of each individual element

78.8=(16+1)×3+M

78.8-51=M

27.8g/mol=M

5 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

What is Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium

hram777 [196]

The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.

Keq is the ratio of the concentration of reactant to the concentration of the product.

The balanced equation is

N₂ + 3H₂ = 2NH₃

The equilibrium constant is $\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}$

The given concentrations of the compounds have been:

Ammonia = 3 M

Nitrogen = 1 M

Hydrogen = 2 M

$\rm \dfrac{9}{1\times 8} = 1.5$

Thus, the correct option is D. Keq = 1.5.

Learn more about Keq

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3 0

2 years ago